Let $M$, $N$ be $A$-modules where $A$ is a commutative ring with $1$. I am studying $M \otimes_{A} N \simeq A^{(M \times N)}/(4 \text{ generators that makes $\otimes$ bilinear})$, as module isomorphism preserving tensor structure. I would like to know any way to identify elements of this structure.

For an easy example, let $M = 2\mathbb{Z}$ and $N = \mathbb{Z}/2$ with $A = \mathbb{Z}$. Why is $2 \otimes 1$ nonzero in $M \otimes N$? The only "correct" answer I could figure out was the fact that we cannot "break down" this element into nonzero elements using bilinearity, but this seems like too much check for such a simple problem like this. I want an answer to this question with a technique that I can use in more general situations!


The key to dealing with tensor products is to think about them in terms of their universal characterization (assume $R$ is commutative)

Let $M$ and $N$ be two $R$-modules and $b:M\times N\to L$ is an $R$-bilinear map. Then, there exists a unique linear map $\widetilde{b}:M\otimes_R N\to L$ such that $\widetilde{b}(m\otimes n)=b(m,n)$.

Now, consider the map $2\mathbb{Z}\times \mathbb{Z}_2\to\mathbb{Z}_2$ defined by $(a,b)\mapsto \frac{a}{2}b$. I'm sure you can verify that this is $\mathbb{Z}$-bilinear and so there exists a linear map $f:2\mathbb{Z}\otimes_\mathbb{Z}\mathbb{Z}_2\to\mathbb{Z}_2$ such that $f(2\otimes 1)=1\cdot 1=1$. Now, if $2\otimes 1$ was zero in $2\mathbb{Z}\otimes_\mathbb{Z}\mathbb{Z}_2$ then every linear map would carry it to zero. The fact that we have found a map that DOESN'T implies that it isn't zero.

  • $\begingroup$ So this solution depends on our ability to find such a bilinear map. Right? $\endgroup$ – user123454321 Feb 13 '13 at 7:49
  • $\begingroup$ @GilYoungCheong Exactly. The ability to find such a linear map is the point, the idea behind, and the correct way to think about tensor products--they are objects that allow us to turn bilinear maps into linear ones. $\endgroup$ – Alex Youcis Feb 13 '13 at 7:50
  • $\begingroup$ Thanks. I will meet more problems before I think about this methodology further. $\endgroup$ – user123454321 Feb 13 '13 at 7:51

In this case, the map $2 : \mathbb{Z} \to M$ (i.e. the "multiplication by 2" map) is an isomorphism, and so

$$ \mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z} / 2 \cong M \otimes_{\mathbb{Z}} N$$

This isomorphism is given by $a \otimes_{\mathbb{Z}} b \mapsto (2a) \otimes_{\mathbb{Z}} b$.

For multiple reasons we know $$ \mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z} / 2 \cong \mathbb{Z} / 2$$ and this isomorphism is given by multiplication; i.e. $a \otimes_{\mathbb{Z}} b \mapsto ab$.

And so if we take $2 \otimes_{\mathbb{Z}} 1 \in M \otimes_{\mathbb{Z}} N$, its preimage under the first isomorphism is $1 \otimes_{\mathbb{Z}} 1 \in \mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}/2$ and its image under the second isomorphism is $1\in \mathbb{Z}/2$.

Therefore, $2 \otimes_{\mathbb{Z}} 1 \in M \otimes_{\mathbb{Z}} N$ is nonzero.

Another is that tensor products respect quotients:

$$ M \otimes_{\mathbb{Z}} N \cong \frac{M \otimes_{\mathbb{Z}} \mathbb{Z}}{M \otimes_{\mathbb{Z}} M}$$

The multiplication maps give isomorphisms $M \otimes_{\mathbb{Z}} \mathbb{Z} \to M$ and $M \otimes_{\mathbb{Z}} M \to 4 \mathbb{Z}$, and thus

$$ M \otimes_{\mathbb{Z}} N \cong \frac{2 \mathbb{Z}}{4 \mathbb{Z}} $$

The image of $2 \otimes_{\mathbb{Z}} 1$ by this map is $2$, which is nonzero.

  • $\begingroup$ Well the first isomorphism you gave harms the structure of the tensor product. For example, note that $2 \otimes 1$ is zero in $\mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}/2$. $\endgroup$ – user123454321 Feb 13 '13 at 7:53
  • $\begingroup$ And the image of that element under my isomorphism is $4 \otimes_{\mathbb{Z}} 1$, which is equal to $2 \otimes_{\mathbb{Z}} 2 = 2 \otimes_{\mathbb{Z}} 0 = 0$ in $2\mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}/2$, so no problem. $\endgroup$ – Hurkyl Feb 13 '13 at 7:54
  • $\begingroup$ Let me go through more stuff in order to make correct reading on this. Thanks for the answer. $\endgroup$ – user123454321 Feb 13 '13 at 7:56

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